If L.C.M of three natural numbers $a, b, c$ is $p^2q^2r^2$(where $p, q, r$ are different prime numbers) such that sum of all possible triplets (a,b,c) are given by $m^n$ (where $m$ and $n$ are prime), then find the value of ($m-4n$)
This is how solved.
As the L.C.M of the given numbers is $p^2q^2r^2$, this means that none of the numbers a,b,c can have $p^0 orp^1,q^0orq^1,r^0 or r^1$ as their factors.
But they can have $p,q,r$ as their factors in powers of two or higher.
For example if $a=p^2q^5r^7$,$b=p^7q^2r^3$ and $c=p^9q^8r^2$ then the L.C.M of a,b,c is $p^2q^2r^2$.
Hence there should be infinite solution i.e. infinite ordered triplets.
But the author has provided with the following solution:
L.C.M. $p^2$ can be obtained from the numbers a, b, c in the following combinations.
$$p^2 p^2 p^2 ⇒ 1 arrangement$$
$$p^2 p^1 p^2 ⇒ 3 arrangement$$
$$p^2 p^1 p^1 ⇒ 1 arrangement$$
$$p^2 p^2 p^0 ⇒ 3 arrangement$$
$$p^2 p^1 p^0 ⇒ 6 arrangement$$
$$p^2 p^0 p^0 ⇒ 3 arrangement$$
$$———————————––—
19 arrangement$$
So if L.C.M. is $p^2q^2r^2$
total combinations are = $19^3$ = $m^n$
⇒ m – 4n = 7.
This is where I am confused and unable to comprehend as why the solution is provided in such a way, because in the above arrangements L.C.M will not be $p^2q^2r^2$.
With your example of $a=p^2q^5r^7$, $b=p^7q^2r^3$ and $c=p^9q^8r^2$, then $\operatorname{lcm}(a,b,c) = p^9 q^8 r^7$. Instead, your stated result is actually $\gcd(a,b,c) = p^2 q^2 r^2$. Remember that $\operatorname{lcm}$ is the least common multiple, i.e., the smallest positive integer which is a multiple of $a$, $b$ and $c$. The value you got of $p^2 q^2 r^2$ is the greatest common divisor, i.e., the largest positive integer which is a divisor of $a$, $b$ and $c$.
With the prime factors decomposition, the exponents of each of the primes in $\gcd(a,b,c)$ is the minimum among those of $a$, $b$ and $c$, while the exponents of each of the primes in $\operatorname{lcm}(a,b,c)$ is the maximum among those of $a$, $b$ and $c$.